--%>

Schrodinger equation with particle in a box problem.

Three dimensional applications of the Schrodinger equation are introduced by the particle-in-a-box problem.

So far only a one-dimensional problem has been solved by application of the Schrodinger equation. Now the allowed energies and the probability functions for a particle that is free to move in three dimensions are deduced. A molecule of a gas enclosed in a cubic container provides a specific example that is dealt with in the section after the general procedure has been developed.

For any three-dimensional problem, the potential energy is, general, a function of three coordinates. For a cubic potential box, the Cartesian coordinates are convenient. The differential equation that must be solved is now the Schrodinger equation in three dimensions.

1310_Particle in a box.png 

For a "cubic box," the potential function can be expressed in terms of separate x, y, and z components,
98_Particle in a box1.png 

Each of the potential function components for a "particle-in-a-box" is like the one-dimensional potential for a "particle-on-a-line".

For three-dimensional systems, the solution function ψ depends on the three coordinates necessary to locate a point in space. It is often profitable to try to separate such systems into parts, with each part involving only one coordinate. On the basis we try the substitution

ψ (x, y, z) = Ø(x)Ø(y)Ø(z)

Substitution of (2) from (1) gives

1350_Particle in a box2.png 

Division by Ø(x)Ø(y)Ø(z) gives

1180_Particle in a box3.png 

For the equation to be satisfied for all values of x, y and z, each term on the left must equal a component of ε, and we can write

ε = εx + εy + εz

The Schrodinger equation can then be broken down into three identical equations of the type

1394_Particle in a box4.png 

Or

578_Particle in a box6.png 

These equations are identical to that written for one-dimensional problem. The solution to the three-dimensional cubic-box problem is therefore

ψ =  Ø(x)Ø(y)Ø(z)

With

1809_Particle in a box7.png

   Related Questions in Chemistry

  • Q : P block why BiF3 is ionic whereas other

    why BiF3 is ionic whereas other trihalides are covalent in nature?

  • Q : Reason for medications contain hcl What

    What is the reason behind this that some medications contain hcl?

  • Q : Molecular mass from Raoults law Provide

    Provide solution of this question. Determination of correct molecular mass from Raoult's law is applicable to: (a) An electrolyte in solution (b) A non-electrolyte in a dilute solution (c) A non-electrolyte in a concentrated solution (d) An electrolyte in a liquid so

  • Q : Electron Spin The total angular

    The total angular momentum of an atom includes an electron spin component as well as an orbital component.The orbital motion of each electron of an atom contributes to the angular momentum of the atom, as described earlier. An additional

  • Q : Dipole attractions for london dispersion

    Illustrate how are dipole attractions London dispersion forces and hydrogen bonding similar?

  • Q : Solubility product On passing H 2 S gas

    On passing H2S gas through a particular solution of Cu+ and Zn+2 ions, first CuS is precipitated because : (a)Solubility product of CuS is equal to the ionic product of ZnS (b) Solubility product of CuS is equal to the solubility product

  • Q : Benefits of soapy detergents over the

    What are the benefits of soapy detergents over the soap less detergents? Briefly state the benefits?

  • Q : Calculating Formulae Superphosphate has

    Superphosphate has the formula CaH4(PO4)2 H2O, what is the calculation to get the percentage of Phosphorus, I need to show the calculation. I know it is 30.9737622 u in weight and 2 atoms of the formula, but not sure how to work the calculation backwards.

  • Q : Molarity of Sodium hydroxide Select the

    Select the right answer of the question. Molarity of 4% NaOH solution is : (a) 0.1M (b) 0.5M (c) 0.01M (d) 0.05M

  • Q : Relationship between free energy and

    The free energy of a gas depends on the pressure that confines the gas. The standard free energies of formation, like those allow predictions to be made of the possibility of a reaction at 25°C for each reagent at